a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)

                \(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)

                \(\dfrac{3}{2}+x=\dfrac{3}{2}\)

                       \(x=\dfrac{3}{2}-\dfrac{3}{2}\)

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)

\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(x=\dfrac{3}{2}+\dfrac{1}{2}\)

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\)                               \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)

\(2x=\dfrac{3}{2}\)                                                     \(2x=\dfrac{1}{2}\)

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\)                hoặc                  \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

\(=>2-3x=\dfrac{3}{2}\)                                               \(2-3x=\dfrac{-3}{2}\)

\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\)                                      \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

Bài 1:

a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)

b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{10}\)

c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)

\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)

\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)

\(=1\)

Bài 2:

a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)

b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)

c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)

d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)

Bài 3:

a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)

\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)

\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)

\(=9.\dfrac{2}{3}.-1\)

\(=6.-1=-6\)

b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)

\(#Wendy.Dang\)

Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))

Võ Ngọc Phương

Bài  3b, kết quả -(3)⁶ = - 729 em nhá chứ không phải (-3)⁶

a: =>2x-1=-2

=>2x=-1

hay x=-1/2

b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)

c: x/8=9/4

nên x/8=18/8

hay x=18

d: \(\Leftrightarrow\left(x-3\right)^2=36\)

=>x-3=6 hoặc x-3=-6

=>x=9 hoặc x=-3

e: =>-1,7x=6,12

hay x=-3,6

h: =>x-3,4=27,6

hay x=31

a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)

\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)

\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)

\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)

\(\dfrac{1}{3}=-2x+1\div6\)

\(x=-\dfrac{1}{2}\)

b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)

\(TH1:3x+2=0\)

\(3x=0-2\)

\(3x=-2\)

\(x=\dfrac{-2}{3}\)

\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)

\(\left(\dfrac{-2}{5}x-7\right)=0\)

\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)

\(\left(\dfrac{-2x-35}{5}\right)=0\)

\(-2x-35=0\)

\(-2x=0+35\)

\(x=-\dfrac{35}{2}\)

c) \(\dfrac{x}{8}=\dfrac{9}{4}\)

\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)

\(x=18\)

d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)

\(x-3=18+2\)

\(x=20-3\)

\(x=17\)

e) \(4,5x-6,2x=6,12\)

\(\dfrac{9x}{2}-6,2.x=6,12\)

\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)

\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)

\(\dfrac{45x-62x}{10}=6.12\)

\(=-17x\div10=6.12\)

\(-17x=10.6.12\)

\(x=-3,6\)

h) \(11,4-\left(x-3,4\right)=-16,2\)

\(x-3,4=-16,2+11,4\)

\(x-3,4=-4,8\)

\(x=-1,4\)

a) (x-1):2/3=-2/5

=>x-1=-4/15

=>x=11/15

b) |x-1/2|-1/3=0

=>|x-1/2|=1/3

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\) 

c) Tương Tự câu B

Bài 2:

a: =>x^2=60

=>\(x=\pm2\sqrt{15}\)

b: =>2^2x+3=2^3x

=>3x=2x+3

=>x=3

c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)

\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)

=>1/2x-2=4

=>1/2x=6

=>x=12

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)